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Statics-Two wires holding traffic light

  1. Nov 19, 2011 #1
    Screen Shot 2011-11-19 at 8.48.14 PM.png

    I dont know what i did wrong

    Find the tension in the two wires supporting the traffic light
    Mass of light= 33 kg
    θ of FT1= 53°
    θ of FT2=37°

    Ʃfx=0
    Ft1cos53-ft2cos37=0
    ft1cos53=ft2cos37

    Ʃfy=mg
    ft1sin53-ft2sin37=mg
    ft1sin53=mg+ft2sin37
    ft1=(ft2sin37+mg)/sin53
    ft1=1.33ft2

    1.33ft2=(ft2sin37+mg)/sin53
    1.33ft2=.72ft2+mg
    .61ft2=323.73 N
    Ft2=530.7n

    ft1cos53=ft2cos37
    ft1cos53=423.84n
    ft1=704.26n
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 19, 2011 #2

    PhanthomJay

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    Tension forces always pull away from the objects on which they act. In the horizontal direction, your equation is correct. Not so in the vertical direction.
     
  4. Dec 8, 2012 #3
    This is what I got:
    TLsin(53o) + TRsin(37o) - 294 N= 0 TRcos(37o) - TLcos(53o) = 0
    TR = TLcos(53o)/cos(37o)

    TLsin(53o) + TRsin(37o) - 294N= 0
    TLsin(53o) + {TLcos(53o)/cos(37o)}sin(37o) - 294 = 0
    TLsin(53o) + TLcos(53o)tan(37o) = 294N
    TL(sin(53o) + cos(53o)tan(37o)) = 294N
    TL = (294 N)/(sin(53o) + cos(53o)tan(37o)) = 234.79884N = 230N

    TR = TLcos(53o)/cos(37o)
    TR = (234.79884 N)cos(53o)/cos(37o) = 176.9336 N = 180N

    Left wire has 230N of tension
    Right wire has 180N of tension
     
  5. Dec 8, 2012 #4
    I don't know where you got the 294N from? 33kg * 9.8 = 323N. g is 9.8, not 8.9!

    Your maths seems right though.

    You can check your answer easily also. 230sin(53) + 180sin(37) = 292N, which is close enough (since you rounded your answers) to 294.
     
  6. Dec 8, 2012 #5
    Well dangit... I guess I got that one wrong then!
     
  7. Dec 8, 2012 #6
    PhantomJay is correct

    jzCpx.jpg

    Hope this helps.
     
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